# Kummer Theory

• Jul 8th 2008, 07:50 PM
ThePerfectHacker
Kummer Theory
Let $K/F$ be a Kummer $n$-extension. Write $K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$. If we define $\text{KUM}(K/F) = \{a\in K^{\times} | a^n \in F^{\times}\}$ and $\text{kum}(K/F) = \text{KUM}(K/F)/F^{\times}$ then it can be shown that $\text{Gal}(K/F) \simeq \text{kum}(K/F)$. Therefore, the degree of the extension, $[K:F]$, is $|\text{kum}(K/F)|$. Define the embedding $\phi: \text{kum}(K/F) \mapsto F^{\times}/(F^{\times})^n$ by $\phi(\alpha F^{\times}) = \alpha^n (F^{\times})^n$. Then, $|\text{im}(\phi)| = |\text{kum}(K/F)| = [K:F]$. It turns out that $\text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$ (this notation means 'group generated by'). By the embedding it means $[K:F] = |\left< a_1(F^{\times})^n,...,a_r(F^{\times})^n \right>|$. And this gives us a way to find the degree of a Kummer extension. What I do not see is why $\text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$. I tried out some examples with $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and convinced myself that the only way to get an element when squared is when the element is a product of the $n$-th roots. But I cannot find a nice proof to this.
• Jul 9th 2008, 06:33 AM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
What I do not see is why $\text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$.

that's a fairly quick result of this theorem (before i explain why, you need to tell me if you basically know this theorem or not):

Theorem: the map $H \mapsto F(H^{\frac{1}{n}})$ is a bijection from the class of all subgroups of $F^{\times}$ which contain $(F^{\times})^n$ onto the class of all Kummer

n-extensions of $F.$
• Jul 9th 2008, 07:50 AM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
that's a fairly quick result of this theorem (before i explain why, you need to tell me if you basically know this theorem or not):

Theorem: the map $H \mapsto F(H^{\frac{1}{n}})$ is a bijection from the class of all subgroups of $F^{\times}$ which contain $(F^{\times})^n$ onto the class of all Kummer

n-extensions of $F.$

No, I am not familar with this theorem. I do not even understand the notation you are using.

However, I did think of an elementary way of proving $\text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$. We note that every element in $K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ has to be expressed in terms of powers and linear combinations of $\sqrt[n]{a_i}$'s. For example, consider $K=F(\sqrt{2},\sqrt{3})$ (here $F=\mathbb{Q}$) then $\{1,\sqrt{2}\}$ is basis for $F(\sqrt{2})/F$ and $\{1,\sqrt{3}\}$ is basis for $F(\sqrt{2},\sqrt{3})/F(\sqrt{2})$. Thus, basis for $K/F$ is $\{1,\sqrt{2},\sqrt{3},\sqrt{2} \cdot \sqrt{3}\}$. Therefore, any element can be written as $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{2}\cdot \sqrt{3}$. And this is true in general, any element in a Kummer extension can be written out in terms of linear combinations of the products of the $n$-th roots. Let $\{x_1,...,x_m\}$ be a basis for $K= F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ where $x_i$'s are products of the $\sqrt[n]{a_j}$'s by above. If $\alpha \in K$ then we can write $\alpha = f(x_1,...,x_m)$ where $f$ is a polynomial over $F$ with $m$ variables. Assume that $\alpha$ has the property that $\alpha^n \in F$. We wish to argue that all terms of $f(x_1,...,x_m)$ must be zero except for possibly one of them, and this will show that the only elements in the Kummer group are the products of the $n$-th roots. Suppose this was not the case and there were two coefficients for $x_i,x_j$ which were non-zero. Then in the expansion of $\alpha^n = f^n (x_1,...,x_m)$ we would have a non-zero "cross term" $x_ix_j$. But then this means $\alpha^n \not \in F$ by the uniqueness of the representation of an element in terms of a basis.
• Jul 9th 2008, 06:29 PM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
No, I am not familar with this theorem. I do not even understand the notation you are using.

However, I did think of an elementary way of proving $\text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$. We note that every element in $K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ has to be expressed in terms of powers and linear combinations of $\sqrt[n]{a_i}$'s. For example, consider $K=F(\sqrt{2},\sqrt{3})$ (here $F=\mathbb{Q}$) then $\{1,\sqrt{2}\}$ is basis for $F(\sqrt{2})/F$ and $\{1,\sqrt{3}\}$ is basis for $F(\sqrt{2},\sqrt{3})/F(\sqrt{2})$. Thus, basis for $K/F$ is $\{1,\sqrt{2},\sqrt{3},\sqrt{2} \cdot \sqrt{3}\}$. Therefore, any element can be written as $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{2}\cdot \sqrt{3}$. And this is true in general, any element in a Kummer extension can be written out in terms of linear combinations of the products of the $n$-th roots. Let $\{x_1,...,x_m\}$ be a basis for $K= F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ where $x_i$'s are products of the $\sqrt[n]{a_j}$'s by above. If $\alpha \in K$ then we can write $\alpha = f(x_1,...,x_m)$ where $f$ is a polynomial over $F$ with $m$ variables. Assume that $\alpha$ has the property that $\alpha^n \in F$. We wish to argue that all terms of $f(x_1,...,x_m)$ must be zero except for possibly one of them, and this will show that the only elements in the Kummer group are the products of the $n$-th roots. Suppose this was not the case and there were two coefficients for $x_i,x_j$ which were non-zero. Then in the expansion of $\alpha^n = f^n (x_1,...,x_m)$ we would have a non-zero "cross term" $x_ix_j$. But then this means $\alpha^n \not \in F$ by the uniqueness of the representation of an element in terms of a basis.

this proof is not complete! you can't just extend things from a simple extension such as $\mathbb{Q}(\sqrt{2},\sqrt{3})$ to the general case! ok, let's see:

i think you chose your basis to be all possible products of powers of $\sqrt[n]{a_j}.$ then you need to show that this set is linearly independent

over F. then when you choose an element of $\alpha$ (which by the way would be just a simple linear combination of $x_j$ and you don't need

to consider it as a polynomial $f(x_1, \ ... \ , x_m)$), then in computing $\alpha^n$ the charactersitic of F will also get involved!
• Jul 9th 2008, 07:01 PM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
this proof is not complete! you can't just extend things from a simple extension such as $\mathbb{Q}(\sqrt{2},\sqrt{3})$ to the general case!

Consider this. Let $K=F(\sqrt[4]{a_1},\sqrt[4]{a_2},\sqrt[4]{a_3})$. Then $\{1,\sqrt[4]{a_1},\sqrt[4]{a_1}^2,\sqrt[4]{a_1}^3\}$ will spam $F(\sqrt[4]{a_1})/F$. It is not necessarily a basis so $\{1,\sqrt[4]{a_1}^2\}$ is a basis - say for simplicity sake. Now $\{1,\sqrt[4]{a_2},\sqrt[4]{a_2}^2,\sqrt[4]{a_3}^3\}$ will spam $F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F(\sqrt[4]{a_1})$. But it is not necessarily a basis. Say $\{1,\sqrt[4]{a_2}\}$ is this basis. Thus, $\{1,\sqrt[4]{a_1}^2,\sqrt[4]{a_2},\sqrt[4]{a_1}^2\sqrt[4]{a_2}\}$ is basis for $F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F$. And so on. The same idea for $K/F$. Thus, any element is a linear combination of the extraction of roots in general.

Quote:

to consider it as a polynomial $f(x_1, \ ... \ , x_m)$), then in computing $\alpha^n$ the charactersitic of F will also get involved!
But the charachteristic of $F$ should not come into play, should it? Because since $F$ has all $n$-roots of unity it means the characheristic of $F$ does not divide $n$. Thus, we do not get multiple of $n$ when we expand $f(x_1,...,x_m)^n$. Do you buy that?
• Jul 9th 2008, 07:48 PM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
Consider this. Let $K=F(\sqrt[4]{a_1},\sqrt[4]{a_2},\sqrt[4]{a_3})$. Then $\{1,\sqrt[4]{a_1},\sqrt[4]{a_1}^2,\sqrt[4]{a_1}^3\}$ will spam $F(\sqrt[4]{a_1})/F$. It is not necessarily a basis so $\{1,\sqrt[4]{a_1}^2\}$ is a basis - say for simplicity sake. Now $\{1,\sqrt[4]{a_2},\sqrt[4]{a_2}^2,\sqrt[4]{a_3}^3\}$ will spam $F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F(\sqrt[4]{a_1})$. But it is not necessarily a basis. Say $\{1,\sqrt[4]{a_2}\}$ is this basis. Thus, $\{1,\sqrt[4]{a_1}^2,\sqrt[4]{a_2},\sqrt[4]{a_1}^2\sqrt[4]{a_2}\}$ is basis for $F(\sqrt[4]{a_1},\sqrt[4]{a_2})/F$. And so on. The same idea for $K/F$. Thus, any element is a linear combination of the extraction of roots in general.

then you need to prove that in general K has a basis chosen from the set of products of powers of $\sqrt[n]{a_j}$ which is

closed under multiplication ... and proving the linear independence in general is not an easy task!

Quote:

But the charachteristic of $F$ should not come into play, should it? Because since $F$ has all $n$-roots of unity it means the characheristic of $F$ does not divide $n$. Thus, we do not get multiple of $n$ when we expand $f(x_1,...,x_m)^n$. Do you buy that?
i know that the char(F) doesn't divide n, but char(F) can still divide some of the coefficients in the expansion of $\alpha^n.$
• Jul 11th 2008, 10:07 AM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
then you need to prove that in general K has a basis chosen from the set of products of powers of $\sqrt[n]{a_j}$ which is

closed under multiplication ... and proving the linear independence in general is not an easy task!

Let $E_2/E_1$ be an algebraic extension and let $\alpha \in E_2$. It is a known result that if $\deg(\alpha,E_2) = l$ then $\{1,\alpha,...,\alpha^{l-1}\}$ is a basis for $E_1(\alpha)/E_1$. Furthermore, if $E_1\subseteq E_2\subseteq E_3$ are fields with $E_3/E_2$ having basis $\{\gamma_i : i \in I\}$ and $E_2/E_1$ having basis $\{\delta_j : j\in J\}$ then $\{ \gamma_i\delta_j\}$ is basis for $E_3/E_1$. Now let us apply these facts to $K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$. Let $\alpha_k = \sqrt[n]{a_k}$. Now define $L_0=F,L_1=F(\alpha_1),L_2=F(\alpha_1,\alpha_2),... ,L_r = F(\alpha_1,...,\alpha_r) = K$. We have the tower of fields $L_0\subseteq L_1\subseteq L_2\subseteq ... \subseteq L_r$. Let $d_i = [L_i : L_{i-1}]$ for $1\leq i\leq n$. Because $x^n - a_i$ has $\alpha_i$ as a root it means $d_i\leq n$. This means $\{1,\alpha_i,...,\alpha_i^{d_i}\}$ is a basis for $L_i/L_{i-1}$. Thus, $B=\{ \alpha_1^{k_1}\alpha_2^{k_2}\cdot ... \cdot \alpha_r^{k_r}| 0\leq k_i\leq d_i \}$ is a basis for $K=L_r/L_0=F$. Let $x_1,...,x_m$ be all the elements of $B$. This means if $\beta \in K$ then we can write $\beta = \Sigma_{j=1}^m c_jx_j$.

If $\beta^n \in L_0$ then all $c_j$'s are zero expect for possibly one of them. We will assume $\text{char}(L_0) =0$. If this was not the case then there are $x_1,x_2$ (by relabeling if necessary) in the linear expansion of $\beta$ such that the coefficients of $x_1,x_2$ are non-zero. Then in the expansion of $(c_1x_1+...+c_mx_m)^n$ we will have a 'cross term' $x_1x_2$ which a non-zero coefficient. Let $x_1 = \alpha_1^{k_1}\cdot ... \cdot \alpha_r^{k_r}$ and $x_2 = \alpha_1^{s_1}\cdot ... \cdot \alpha_r^{s_r}$ where $0\leq k_j,s_j\leq d_j$. Then the cross term $x_1x_2 = \alpha_1^{k_1+s_1}\cdot ... \cdot \alpha_r^{k_r+s_r}$. Now realize that $0\leq k_j+s_j\leq 2d_j$. For this to be an element of $L_0$ we need that $k_j+s_j$ are multiples of $d_j$. Note that if $k_j+s_j>n$ then we can write $\alpha_j^{k_j+s_j} = \alpha_j^n \alpha_j^{k_j+s_j-n}$ with $\alpha_j^n$. Because of the constrainst it is impossible for all $k_j+s_j$ to be multiples of $d_j$ unless $x_1=x_2$ which is not the case. It seems that the problem is that that we do not know if there is a triple cross term, for example, $x_1x_2x_3$ whose basic reduced form (of the exponents) is not the same as $x_1x_2$. And therefore it is possible we get cancellation from the cross terms.
• Jul 15th 2008, 02:07 PM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
Theorem: the map $H \mapsto F(H^{\frac{1}{n}})$ is a bijection from the class of all subgroups of $F^{\times}$ which contain $(F^{\times})^n$ onto the class of all Kummer

How do you prove the generating fact using this theorem?
• Jul 16th 2008, 08:42 PM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
How do you prove the generating fact using this theorem?

let $H_1=(K^{\times})^n \cap F^{\times},$ and $H_2=<(F^{\times})^n,a_1,a_2, \ ... \ , a_r>.$ then both $H_1$ and $H_2$ are subgroups of $F^{\times}$ which contain $(F^{\times})^n.$

Claim: $F(H_1^{\frac{1}{n}})=F(H_2^{\frac{1}{n}})=K.$

Proof: since $\sqrt[n]{a_j} \in H_1^{\frac{1}{n}},$ and $\sqrt[n]{a_j} \in H_2^{\frac{1}{n}},$ we must have $K \subseteq F(H_1^{\frac{1}{n}}),$ and $K \subseteq F(H_2^{\frac{1}{n}}).$ now if $x \in H_1^{\frac{1}{n}},$ then $x^n =\alpha^n,$ $\alpha \in K^{\times}.$

thus $x=\alpha \xi,$ for some n-th root of unity $\xi.$ hence $x \in K.$ so $F(H_1^{\frac{1}{n}}) \subseteq K.$ finally if $x \in H_2^{\frac{1}{n}},$ then $x^n=u^n a_1^{t_1}a_2^{t_2} \ ... \ a_r^{t_r},$ for some

$u \in F^{\times},$ which gives us $x=u\sqrt[n]{a_1^{t_1}} \sqrt[n]{a_2^{t_2}} \ ... \ \sqrt[n]{a_r^{t_r}} \xi,$ for some n-th root of unity $\xi.$ so $x \in K,$ and thus $F(H_2^{\frac{1}{n}}) \subseteq K. \ \ \ \ \square$

so by the theorem $H_1=H_2.$ now it's obvious that $\text{kum}(K/F)=<\sqrt[n]{a_1}F^{\times}, \sqrt[n]{a_2}F^{\times}, \ ... \ , \sqrt[n]{a_r}F^{\times}>.$
• Jul 16th 2008, 09:14 PM
ThePerfectHacker
Are you defining $H_1^{1/n} = \{ x\in H_1 | x^n \in H_1 \}$?
I just never seen that notation before.

EDIT: Yes it is. Sorry, I was going to delete this but I see you are replying.
• Jul 16th 2008, 09:27 PM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
Are you defining $H_1^{1/n} = \{ x\in H_1 | x^n \in H_1 \}$?
I just never seen that notation before.

$x \in H_1??$ no! $x \in H_1^{\frac{1}{n}},$ if, in some extension of $H_1$, it's a root of $t^n - \alpha = 0,$ for some $\alpha \in H_1.$
• Jul 17th 2008, 09:46 AM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
Theorem: the map $H \mapsto F(H^{\frac{1}{n}})$ is a bijection from the class of all subgroups of $F^{\times}$ which contain $(F^{\times})^n$ onto the class of all Kummer n-extensions of $F.$

I would restate the theorem as: $H\mapsto F(H^{1/n})$ is a bijection from the set of all subgroups of $F^{\times}$ which contain $(F^{\times})^n$ onto the set of all Kummer n-extension of $F$ which are contained in $\bar F$ (the algebraic closure of $F$). Is this okay (this does not damage the theorem in any way) ?

Quote:

Originally Posted by NonCommAlg
$x \in H_1??$ no! $x \in H_1^{\frac{1}{n}},$ if, in some extension of $H_1$, it's a root of $t^n - \alpha = 0,$ for some $\alpha \in H_1.$

That was stupid of me. When I looked at your proof I used $\{x\in \bar F | x^n \in H_1\}$.

Quote:

Originally Posted by NonCommAlg
let $H_1=(K^{\times})^n \cap F^{\times},$ and $H_2=<(F^{\times})^n,a_1,a_2, \ ... \ , a_r>.$ then both $H_1$ and $H_2$ are subgroups of $F^{\times}$ which contain $(F^{\times})^n.$

Claim: $F(H_1^{\frac{1}{n}})=F(H_2^{\frac{1}{n}})=K.$

Proof: since $\sqrt[n]{a_j} \in H_1^{\frac{1}{n}},$ and $\sqrt[n]{a_j} \in H_2^{\frac{1}{n}},$ we must have $K \subseteq F(H_1^{\frac{1}{n}}),$ and $K \subseteq F(H_2^{\frac{1}{n}}).$ now if $x \in H_1^{\frac{1}{n}},$ then $x^n =\alpha^n,$ $\alpha \in K^{\times}.$

thus $x=\alpha \xi,$ for some n-th root of unity $\xi.$ hence $x \in K.$ so $F(H_1^{\frac{1}{n}}) \subseteq K.$ finally if $x \in H_2^{\frac{1}{n}},$ then $x^n=u^n a_1^{t_1}a_2^{t_2} \ ... \ a_r^{t_r},$ for some

$u \in F^{\times},$ which gives us $x=u\sqrt[n]{a_1^{t_1}} \sqrt[n]{a_2^{t_2}} \ ... \ \sqrt[n]{a_r^{t_r}} \xi,$ for some n-th root of unity $\xi.$ so $x \in K,$ and thus $F(H_2^{\frac{1}{n}}) \subseteq K. \ \ \ \ \square$

so by the theorem $H_1=H_2.$ now it's obvious that $\text{kum}(K/F)=<\sqrt[n]{a_1}F^{\times}, \sqrt[n]{a_2}F^{\times}, \ ... \ , \sqrt[n]{a_r}F^{\times}>.$

I think I understand your proof if I do it in the following manner: Given $K$ and $F$. Construct $\bar F$ so that $K\subset \bar F$. Define $H_1^{1/n} = \{x\in \bar F | x^n \in H_1\}$ and $H_2^{1/n} = \{ x\in \bar F | x^n \in H_2\}$. So this is basically what you are doing except I picked a big extension field to work under - this makes me feel safe. And now I just carry through the exact same argument. I think that is okay to do.

My last question is how do you prove this bijection theorem between subgroups and Kummer n-extensions. In fact, I do not even see why $F(H^{1/n})$ is a Kummer n-extension over $F$. The definition I am using is that $K/F$ is a Kummer n-extension when $K/F$ is Galois and $F$ has primitive $n$-th root of unity and $\text{Gal}(K/F)$ is a finite abelian group whose exponent divides $n$. I can see that $F(H^{1/n})$ is a splitting field over $\{ x^n - h | h\in H\}$ and each polynomial is seperable over $F$. However, I do not see why $\text{Gal}(F(H^{1/n})/F)$ is a finite abelian group whose exponent divides $n$.
Note, I am familar with the theorem that $B: \text{Gal}(K/F)\times \text{kum}(K/F)\to \{1,\zeta,...,\zeta^{n-1}\}$ defined by $B(\sigma, \alpha F^{\times}) = \sigma(\alpha)/\alpha$ is a non-degenerate pairing. And so $\text{kum}(K/F) \simeq \text{hom}(\text{Gal}(K/F), \{1,...,\zeta^{n-1}\})$. Maybe your bijection theorem is somehow related to it, but I do not see the connection.

We know that $((F^{\times})^{1/n} \cap K^{\times})/F^{\times}$ (the Kummer group) is isomorphic to $\text{Gal}(K/F)$. Let $H = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times} \right>$ and $U$ be the $n$-th roots of unity. Define $B: H\times \text{Gal}(K/F)\to U$ by $B(\alpha F^{\times},\sigma) = \sigma(\alpha)/\alpha$. This is a bilinear pairing, since it is non-degenerate it means $H\simeq \text{hom}(\text{Gal}(K/F),U)$. But $\text{hom}(\text{Gal}(K/F),U)\simeq \text{hom}(\text{Gal}(K/F),\mathbb{C})\simeq \text{Gal}(K/F)$ since $\text{Gal}(K/F)$ is a finite abelian group it is isomorphic to its charachter group. And this means that $H$ is isomorphic to the Kummer group, since $H$ is a subgroup of the Kummer group it means that $H$ in fact the Kummer group.